Proof for intersection.

Question

Really stuck on this one, seemed simple but I just couldn't get it for some reason.

Suppose that $\mathcal U$ is the universal set, and that $A$, $B$ and $C$ are three arbitrary sets of elements of $\mathcal U$. Prove that if $C \setminus A = B$, then the intersection of $A$ and $B$ is empty. Hint: use an indirect proof.

Answer 1

Show $C$ \ $A=B$ implies $A \cap B=\emptyset$.

Assume $A\cap B \not = \emptyset.$

Then there is a $x$ s.t $x \in A$ and $x \in B$.

Since $x \in B$, and $C$ \ $A =B$ , we have $x \in C$\ $A,$

then $x \not \in A$, a contradiction.

Answer 2

The following is a direct proof:

$1$. Assume $C-A = B$

$2$. Then, by substitution,

$$A \cap B = A \cap \big(C-A \big)$$

$3$. By definition of set difference,

$$=A \cap \big(C \cap A^c\big)$$

$4$. By commutative law,

$$=A \cap \big(A^c \cap C\big)$$

$5$. By associative law,

$$=\big(A \cap A^c\big) \cap C$$

$6$. By negation law,

$$=\emptyset \cap C$$

$7$. By domination law,

$$=\emptyset$$

Therefore, if $C-A = B$, then $A \cap B = \emptyset$